This work W = U I t. must be supplied by the system. In order to check whether a current density (J) of 10exp11 A / m2 is sufficient, we use for the superconducting layer
W = U I t = U x (J x F) x s / v => J = W x v / U x F x s

I thought you might find this interesting Gerhard. A few weeks after the F&P announcement Larry A Hull (in a letter to Chemical and Engineering News, May 15, 1989, page 3) proposed the mechanism of a deuteron capturing an electron to become nn, “dineutronium”.

Here is the DOI:
10.1021/cen-v067n020.p002
The letter is available on sci-hub. Nice read. Shows how people reacted a few weeks after F&P announcement.

Alan DeAngelis

Yeah, and there’s a letter from a crackpot named DeAngelis just above Hull’s letter.

Alan DeAngelis

Now (27 years later) I’m thinking that the palladium of palladium hydride, D2Pd during symmetric IR stretching could fuse with its deuterons to form cadmium in an excited state, Cd* that in turn would fission back into palladium and helium.

Don’t we have to assume Cooper-pairs here instead of ‘classical’ conduction electrons? If so, which consequences would you see both with regard to the energy aspect and the supposed mechanism of electron capture?

hunfgerh

Yes and no.
The quantum-mechanical calculation of the current density to the one performed
here will less differ. However, the current density is only a measure for the
number of charge carriers.

But the mechanism of “e-capture” can only be carried out if the cooper-pair
(spin 0) decouples after a certain coherent length 10exp-6m -10exp-10m into two
uncoupeld electrons with the (spin ½) in the same direction. “Superconducting
is a periodic decoupl/ coupel process.” For this reason, the same number of unpaired electrons are present in a superconductor in addition to the cooper pairs. Only by this can the s-electronbe pushed into the core by repulsion of similar charges.